Answer:
Part a)
[tex]I = 1.5 kg m^2[/tex]
Part b)
[tex]I = 0.75 kg m^2[/tex]
Part c)
[tex]I = 1.5 kg m^2[/tex]
Explanation:
Part a)
Moment of inertia of the system about an axis passing through B and C is given as
[tex]I = mL^2 + mL^2 + m(0) + m(0)[/tex]
[tex]I = 2mL^2[/tex]
[tex]I = 2(3 kg)(0.50^2)[/tex]
[tex]I = 1.5 kg m^2[/tex]
Part b)
Moment of inertia of the system about an axis passing through A and C is given as
[tex]I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2[/tex]
[tex]I = 2m\frac{L^2}{2}[/tex]
[tex]I = (3 kg)(0.50^2)[/tex]
[tex]I = 0.75 kg m^2[/tex]
Part c)
Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square
[tex]I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2[/tex]
[tex]I = 4m\frac{L^2}{2}[/tex]
[tex]I = 2(3 kg)(0.50^2)[/tex]
[tex]I = 1.5 kg m^2[/tex]